Extending the investigation.

Mathematical Portfolio Higher Level

Patterns within systems of linear equations.

Candidate name: Naumova Roxana.

Table of contents:

Introduction ……………………………………………………… 3

Part A.

Examination of constants …………………………………………………………4

Solving the system ………………………………………………………………..4

Examples of the similar systems ……………………………………………........4

Conjecture & Proof ………………………………………………………….........5

Extending the investigation …………………………………………………….....6

Conjecture for 3×3 equation system ……………………………………………...7

Part B.

Examination of constants ………………………………………………………….9

The graph of the family of the similar equations ………………………………….9

Examination of the graph ………………………………………………………….10

General equation …………………………………………………………………..10

Proof for the graphical pattern …………………………………………………….10

Conclusion ………………………………………………………..11

Introduction.

The knowledge of how different patterns in the linear equation system influence the solution or the graph of the function can help a lot to solve the systems of this specific type.

In this portfolio I am going to investigate the patterns formed by the constants in the linear 2×2 equation systems. I will look at the way those patterns influence the solution of the system and how it basically could be explained.

I am going to display the graphs of these equations and find some graphical patterns and then explain it by providing the reference to the non-graphical method.

I will also try to extend my investigation to 3×3 linear system and see how the solutions are similar or differ from the previous case.

Part A.

Examination of constants of given equations.

x + 2y = 3

2x – y = -4

In the first equation the constants form an arithmetic sequence, where common difference is equaled to 1,U₁ = 1, U₂ = 2, U₃ = 3.

In the second equation the constants form an arithmetic sequence as well. The common difference is equaled to -3 and U₁ = 2, U₂ = -1, U₃ = -4.

In both equations the constants form arithmetic sequences, so we could determine that there is a pattern which could in some way influence the solutions in the similar systems.

Solving the system.

To solve the system given I am going to use the substitution method.

x + 2y = 3

2x – y = -4

x = 3 – 2y

6 – 4y – y = -4

x = 3 – 2y

-5y = -10

x = -1

y = 2

The significance of the solution is that x and y are equaled to the 2nd constants of the given equations.

3. Examples of the similar systems:

a) As for the first example I will create a system of two equations, which constants form an arithmetic sequences with the common differences equaled to “-5” and “1”. To solve it I will use the same substitution method again.

y – 4x = -9

2y + 3x = 4

y = -9 + 4x

11x = 22

y = -1

x = 2

I switched the order of x and y in the original equations, and the solutions also switched.

b) Another example with the common differences “-4” and “1”.

2y – 2x = -6

3y + 4x = 5

y = -1

x = 2

The order on x and y is the same as in the previous equation, the solutions are the same.

c) Next example will contain arithmetic sequences with the common differences “8” and “-6”.

4x + 12y = 20

x – 5y = -11

y = 2

x = -1

If we switch the order of x and y back again, we will get the same solutions as in the given example.

d) Just to be completely confident before making the conjecture, I will check another example of the similar system containing the differences “3” and “5”.

x + 4y = 7

x + 6y = 11

y = 2

x = -1

The answers are the same as in the all the previous cases.

Now after checking all the significant moments I can make a conjecture.

4. Conjecture & Proof.

Considering all the examples above we can make a conjecture:

If the constants in the linear 2×2 equations create an arithmetic sequences and the order of x and y in the equations is the same, then the solution for the first unknown is “-1” and for the second one is “2”.

Proof:

To prove this conjecture I will solve the equation system I general way. To do this I will consider 1st term of the 1st equation being equaled to “n” and the common difference “d”, and the first term of the second equation will be equaled to “k” with the common difference “h”.

nx + (n + d) × y = n + 2d × (k + h)

kx + (k + h) × y = k + 2h × (n + d)

The necessary condition here to precise is that (k + h) ≠ 0

(n + d) ≠ 0

n(k + h)x + (n + d)(k + h)y = (n + 2d)(k + h)

k(n + d)x + (n + d)(k + h)y = (k + 2h)(n + d)

Then we subtract the bottom equation from the top one. The variable “y” disappears meaning we are isolating the equation from the variable.

Now I am going to alienate the following equation from the system and find the value of “x”.

n (k + h) x – k(n + d)x = (n+ 2d)(k + h) – (k + 2h)(n + d)

nkx + nhx – knx – kdx = nk + nh + 2kd + 2dh – kn – kd – 2hn – 2dh

nhx – kdx = - nh + kd

x(nh – kd) = -(nh – kd)

Here we precise that nh – kd ≠ 0, and after we get x = -1.

Now, as we have already found the value of “x”, we go back to the system to find “y”.

x = -1

-n + (n + d) y = n + 2d

x = -1

y (n + d) = 2n + 2d

x = -1

y = 2

The conjecture is proved, which means that if the constants in the linear 2×2 equations create an arithmetic sequences and the order of x and y in the equations is the same, then the solution for the first unknown is “-1” and for the second one is “2”.

Extending the investigation.

Now I am going to make 3×3 linear system of the equations to explore the solutions in case if the constants will also form arithmetic progressions.

First of all I am going to create a random system to check the solutions.

a) I will create 3×3 linear system with common differences “1”, “2” and “-1”.

x + 2y + 3z = 4

-x + y + 3z = 5

3x + 2y + z = 0

To solve this equation we will use matrixes.

A =

Using GDC we will work out the determinant of A, and it appeared to be equaled to 0,which means that the equation either has n infinite number of solutions or no solutions at all.

If we look at these three equations, we can say that each of them is a Cartesian equation of a plane. The intersection of three planes could be either a point, or a line, or a plane. However, considering the fact that the determinant is equaled to 0, there is no finite number of solutions, so, the point does not suit. If the intersection is a plane, then all the three planes should have the same equation which is also not true. The last option is that the solution could be represented as a line in a 3D space.

We can simplify our system: first, let us sum the first and the second equation.

3y + 6z = 9

3x + 2y + z = 0

Now, we will multiply the last equation by 6 and subtract the first one from it.

3y + 6z = 9

18x + 9y = -9

y + 2z = 3

2x + y = -1

In the Cartesian form the equation of this line could be represented as = =

b) The second example will contain the equations with the common differences “-2”, ‘1”, “-5”.

8x + 6y + 4z = 2

x + 2y + 3z = 4

3x – 2y – 7z = -12

Using GDC again the determinant of a matrix is equaled to 0, which means either infinite number of solutions or no solutions at all. However, we will try to work out the equation for a line which represents the infinite number of solutions in this case.

First, we will multiply the second equation by 3 and then subtract it from the last one.

8x + 6y + 4z = 2

-8y – 16z = -24

8x + 6y + 4z = 2

-2y – 4z = -6

And now when we sum the two equations and make simplifications, we get:

2x + y = -1

y + 2z = 3

This is the same equation of a line as in the previous case, which obviously is going to have the same Cartesian form.

c) The common differences this time will be equaled to “7”, “-1” and “-5”.

3x + 10y + 17z = 24

-2x – 3y – 4z = -5

45x + 40y + 35z = 3

Determinant of matrix is 0 again.

After rearranging these equations we finally arrive to the same result as in the previous two cases.

2x + y = -1

y + 2z = 3

d) The last example with the common differences “5”, “-1” and “9”.

5x + 10y + 15z = 20

3x + 2y + z = 0

4x + 13y + 22z = 31

And the determinant is 0 again, the system can be represented as following:

2x + y = -1

y + 2z = 3

Considering the previous conjecture about 2×2 linear system, where all the solutions were the same not depending on the constants and created a point on the graph, we can suggest that for 3×3 linear equation systems the solution for each of them will be represented as the same line with an equation = = . Finally, I am ready to make a conjecture regarding 3 linear system.

6. Conjecture for 3×3 equation system.

Conjecture:

If the constants in linear 3×3 equation system create arithmetic sequences the system will have the infinite number of solutions obeying the following the relation 2x + y = -1

y + 2z = 3,

or represent the line with an equation = = in space.

Proof:

First of all I am going to rewrite my equations in general way as the previous time.

nx + (n + d)×y + (n + 2d)×z = n + 3d

kx + (k + h)×y + (k + 2h)×z = k + 3h

mx + (m + p)×y + (m + 2p)×z = m + 3p

n×(x + y + z) + d×(y + 2z) = n + 3d

k×(x + y + z) + h×(y + 2z) = k + 3h

m×(x + y + z) + p×(y + 2z) = m + 3p

Supposing that x + y + z = u and y + 2z = v, I will rewrite my system.

n×u + d×v = n + 3d

k×u + h×v = k + 3h

m×u + p×v = m + 3p

n×(u – 1) + d×(v – 3) = 0

k×(u – 1) + h×(v – 3)= 0

m×(u – 1) + p×(v – 3) = 0

n×u* + d×v* = 0

k×u* + h×v* = 0

m×u* + p×v* = 0

This type of system is called homogeneous system. It either has the only solution when u* = 0 and v* = 0, or an infinite number of solutions.

u* = 0

v* = 0

u – 1 = 0

v – 3 = 0

x + y + z = 1

y + 2z = 3

x + y + - = 1

z = -

2x + y = -1

y + 2z = 3

The conjecture is proved.

So, if the constants in linear 3×3 equation system create arithmetic sequences the system will have the infinite number of solutions obeying the following the relation 2x + y = -1

y + 2z = 3,

or represent the line with an equation = = in space.

Expending the investigation even more, in 4D for example, we could suggest that all the solutions will belong to one plane.

Part B.

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